I tried converting everything in terms of sinx and cosx but that doesn't seem to help.

Once you've simplified and gotten an expression in just sine, you can use the double-angle formula (in reverse) for the cosine:

. . . . .

Then the integral is pretty straightforward.

Note: You will be using the double-angle formulas (in reverse) a lot in calculus, so make sure you're solid on them!

I'm still not sure I understand. Once I convert this into terms of sinx and cosx I have integral((sinx)^2 / (cosx)^4)dx. After that I should use the double angle forumual or half angle? The one you showed me is the half angle formual.